A pole of length 138 cm stands in a pool of water, such that the top of the pole is 49 cm above the water surface. Sunlight is incident on the pole at an angle of 38 degrees from the vertical. What is the length of the shadow of the pole at the bottom of the pool? Use a refractive index of 1.33 for water and 1 for air. Put your answer in cm and give units.

To solve this problem, we need to use Snell's law of refraction, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of light in the two media, or equivalently, to the opposite of the ratio of the indices of refraction.

1. First, we need to find the angle of refraction when the light enters the water. We can use Snell's law for this:

   sin(i) / sin(r) = n2 / n1

   where i is the angle of incidence, r is the angle of refraction, n1 is the refractive index of the first medium (air in this case), and n2 is the refractive index of the second medium (water in this case).

   The angle of incidence is the complement of the given angle (90 - 38 = 52 degrees). Plugging these values into Snell's law gives us:

   sin(52) / sin(r) = 1.33 / 1

   Solving for r gives us an angle of refraction of approximately 40.5 degrees.

2. Next, we need to find the length of the shadow in the water. We can use the tangent of the angle of refraction for this:

   tan(r) = shadow length / pole length in water

   The pole length in water is the total length of the pole minus the length above water (138 - 49 = 89 cm). Plugging these values into the equation gives us:

   tan(40.5) = shadow length / 89 cm

   Solving for the shadow length gives us approximately 75 cm.

So, the length of the shadow of the pole at the bottom of the pool is approximately 75 cm.